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adelina 88 [10]
2 years ago
9

If your cell phone rings while you are driving and you do not have a hands-free device you should

Computers and Technology
1 answer:
Dmitry_Shevchenko [17]2 years ago
3 0

Answer:

wait to get to a red light or a stop sign then answer it and put it on speaker phone

Explanation:

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What is the easiest way to deploy a C++ program to a user’s computer? a. Copy the source code to the user’s computer and then co
Morgarella [4.7K]

Answer:

c. Copy the executable file to the user’s computer

Explanation:

When you run a c++ program, an executable file (.exe) is created in the same directory with the source file.

To deploy this program on another system, you need to transfer the executable file from the original system where the program was executed to the new system.

The program will work fine without errors.

However, do not mistake the executable file for the source file. The source file ends in .cpp.

8 0
2 years ago
Write a program in Python that reads in investment amount, annual interest rate, and number of years, and displays the future in
pickupchik [31]

Answer:

investmentAmount = float(input("enter the investment amount: "))

annualInterestRate = float(input("enter the Annual Interest Rate: "))

numYears = int(input("Enter NUmber of Years: "))

monthlyInterestRate = annualInterestRate/12

futureInvestmentValue = investmentAmount * (1 + monthlyInterestRate)*(numYears*12)

print("The Future Investment Value is: ")

print(futureInvestmentValue)

Explanation:

Using python programming language as required, we use the input function to prompt user for inputs for each of the variables.

There is a conversion from the variable annualInterestRate  to monthlyInterestRate before the formula for the futureInvestmentValue is applied

4 0
3 years ago
When reading data across the network (i.e. from a URL) in Python 3, what string method must be used to convert it to the interna
EastWind [94]

Answer:

decode( )

Explanation:

We can use the method decode( ) to decode the string using the codec registered for encoding.

There are two parameters

encoding: We can encode with this parameter.

errors: If used to manage the errors.

For example:

String = "this is string example....wow!!!";

String = Str.encode('base64','strict');

If we print these variables we have:

String = b'dGhpcyBpcyBzdHJpbmcgZXhhbXBsZS4uLi53b3chISE='

String = this is string example....wow!!!

7 0
2 years ago
2-Write test programs in java to determine the scope of a variable declared in a for statement. Specifically, the code must dete
ankoles [38]

Answer:

See Explaination

Explanation:

public class testscope

{

//start of main function

public static void main(String[] args)

{

//varible declration

int i;

int x;

//loop for 10 times

for(i=0; i<10; i++)

{

//initialize value of x to 10

x = 10;

}

//the scope of variable x is visible outside of for loop

System.out.println("The value of x is: "+x);

}

}

See attachment for sample output

nb:

You can clearly see in the output of Java program the value of x is not printed and program return errors. It means the variable x declared inside for loop does not has scope outside the for loop.

7 0
3 years ago
. (a) Prove or disprove carefully and in detail: (i) Θ is transitive and (ii) ω is transitive. (b) Assume n is a positive intege
Sergio [31]

Answer:

The Following are the solution to this question:

Explanation:

In Option a:

In the point (i) \Omega is transitive, which means it converts one action to others object because if \Omega(f(n))=g(n) indicates c.g(n). It's true by definition, that becomes valid. But if \Omega(g(n))=h(n), which implies c.h(n). it's a very essential component. If c.h(n) < = g(n) = f(n) \. They  \Omega(f(n))   will also be h(n).  

In point (ii), The  value of \Theta is convergent since the \Theta(g(n))=f(n). It means they should be dual a and b constant variable, therefore a.g(n) could only be valid for the constant variable, that is  \frac{1}{a}\ \  and\ \ \frac{1}{b}.

In Option b:

In this algorithm, the input size value is equal to 1 object, and the value of  A is a polynomial-time complexity, which is similar to its outcome that is O(n^{2}). It is the outside there will be a loop(i) for n iterations, that is also encoded inside it, the for loop(j), which would be a loop(n^{2}). All internal loops operate on a total number of N^{2} generations and therefore the final time complexity is O(n^{2}).

6 0
3 years ago
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